*Initial Order Deformation Problem*

The deformation equations for the initial order can be viewed as follows:

$$(1 - \Phi)L\_{\psi\_1}[\widehat{\psi}\_1(\varepsilon, \Phi) - \widehat{\psi}\_{1,0}(\varepsilon, \Phi)] = \Phi \hbar\_{\psi\_1} H\_{\psi\_1} N\_{\psi\_1} [\widehat{\psi}\_1(\varepsilon, \Phi)] \tag{30}$$

$$(1 - \Phi)L\_{\psi\_2}[\widehat{\psi}\_2(\varepsilon, \Phi) - \widehat{\psi}\_{2,0}(\varepsilon, \Phi)] = \Phi \hbar\_{\psi\_2} H\_{\psi\_2} N\_{\psi\_2} [\widehat{\psi}\_2(\varepsilon, \Phi)] \tag{31}$$

$$(1 - \Phi)L\_{\psi\_3}[\widehat{\psi}\_3(\varepsilon, \Phi) - \widehat{\psi}\_{3,0}(\varepsilon, \Phi)] = \Phi \hbar\_{\psi\_3} H\_{\psi\_3} N\_{\psi\_3} [\widehat{\psi}\_3(\varepsilon, \Phi)] \tag{32}$$

$$\mathbb{E}(1-\Phi)L\_{\psi4}\left[\widehat{\psi}\_{4}(\varepsilon,\Phi)-\widehat{\psi}\_{4,0}(\varepsilon,\Phi)\right]=\Phi\hbar\_{\psi4}H\_{\psi4}N\_{\psi4}\left[\widehat{\psi}\_{4}(\varepsilon,\Phi)\right] \tag{33}$$

$$(1 - \Phi)L\_{\psi\_5}[\widehat{\psi}\_5(\varepsilon, \Phi) - \widehat{\psi}\_{5,0}(\varepsilon, \Phi)] = \Phi \hbar\_{\psi\_5} H\_{\psi\_5} N\_{\psi\_5} [\widehat{\psi}\_5(\varepsilon, \Phi)] \tag{34}$$

and the boundary conditions are

$$
\widehat{\psi}\_1(1,\Phi) = -\mu\_2 \widehat{\psi}\_1'(1,\Phi), \\
\widehat{\psi}\_1(0,\Phi) - 1 = -\mu\_1 \widehat{\psi}\_1'(0,\Phi) \tag{35}
$$

$$
\widehat{\psi\varphi\_2'}(0,\Phi) - 1 = -\mu\_1 \widehat{\psi\varphi\_2'}(0,\Phi), \\
\widehat{\psi\varphi\_2}(1,\Phi) = -\mu\_2 \widehat{\psi\varphi\_2'}(1,\Phi) \tag{36}
$$

$$
\widehat{\psi\_3}^{'}(0,\Phi) = \mu\_1 \widehat{\psi\_3}^{'/'}(0,\Phi), \\
\widehat{\psi\_3}(0,\Phi) = 0, \\
\widehat{\psi\_3}(1,\Phi) = 1, \\
\widehat{\psi\_3}^{'}(1,\Phi) = -\mu\_2 \widehat{\psi\_3}^{'/'}(1,\Phi) \tag{37}
$$

$$
\widehat{\psi}\_4^{'}(0,\Phi) = \mu\_1 \widehat{\psi}\_4^{'/'}(0,\Phi), \\
\widehat{\psi}\_4(0,\Phi) = 0, \\
\widehat{\psi}\_4(1,\Phi) = 1/2, \\
\widehat{\psi}\_4^{'}(1,\Phi) = -\mu\_2 \widehat{\psi}\_4^{'/'}(1,\Phi), \tag{38}
$$

$$
\widehat{\psi}\_5(1,\Phi) = -\mu\_2 \widehat{\psi}\_5^/(1,\Phi), \widehat{\psi}\_5(0,\Phi) - 1 = -\mu\_1 \widehat{\psi}\_5^/(0,\Phi) \tag{39}
$$

where *<sup>N</sup>*ψ1 , *<sup>N</sup>*ψ2 , *<sup>N</sup>*ψ3 , *<sup>N</sup>*ψ4 , and *<sup>N</sup>*ψ5 are defined as

$$N\_{\psi\_3}[\widehat{\psi}\_3(\varepsilon;\Phi)] = \widehat{\psi}\_3^{\prime//\prime/\prime} - R\left(\widehat{\psi}\_3^{\prime}\widehat{\psi}\_3^{\prime\prime/\prime} - \widehat{\psi}\_3\widehat{\psi}\_3^{\prime\prime/\prime}\right) - M^2\psi\_3^{\prime} \tag{40}$$

$$N\_{\widehat{\psi}\_1} \left[ \widehat{\psi}\_1 (\zeta; \Phi) \right] = \widehat{\psi}\_1 \,' \,' - R \left( \widehat{\psi}\_1 \widehat{\psi}\_3 \,' - \widehat{\psi}\_3 \widehat{\psi}\_1 \,' \right) - M^2 \psi\_{1\prime} \tag{41}$$

$$N\_{\widehat{\psi}\_2} \left[ \widehat{\psi}\_2(\boldsymbol{\varepsilon}; \boldsymbol{\Phi}) \right] = \widehat{\psi}\_2 \, ^{\prime \, \prime} + R \Big( \widehat{\psi}\_3 \widehat{\psi}\_2 \, ^{\prime} \Big) - M^2 \psi\_{2^{\prime}} \tag{42}$$

$$N\_{\psi\_4}[\widehat{\psi}\_4(\zeta;\Phi)] = \widehat{\psi}\_4^{\prime//\prime/\prime} - 2\widehat{\kappa}\widehat{\psi}\_4\widehat{\psi}\_4^{\prime//\prime} - M^2\widehat{\psi}\_4^{\prime} \tag{43}$$

$$N\psi\_5\widehat{\left[\psi\_5(\zeta;\Phi)\right]} = \widehat{\psi}\_5^{'/'} - R\left(\widehat{\psi}\_5\widehat{\psi}\_4{'}^{'} - 2\widehat{\psi}\_4\widehat{\psi}\_5{'}^{'}\right) - M^2\widehat{\psi}\_5\tag{44}$$

Here, the auxiliary parameters are ψ1 0, ψ2 0, ψ3 0, ψ4 0, and ψ5 0, while the non-zero auxiliary functions are expressed as *<sup>H</sup>*ψ1 , *<sup>H</sup>*ψ2 , *<sup>H</sup>*ψ3 , *<sup>H</sup>*ψ4 , and *<sup>H</sup>*ψ5 , and ς ∈ [0, 1] is the embedding parameter.

From Equations (30)–(34), it is observed that when ς = 0 there is

$$
\widehat{\psi}\_1(\varepsilon,0) = \widehat{\psi}\_{1,0}(\varepsilon), \widehat{\psi}\_2(\varepsilon,0) = \widehat{\psi}\_{2,0}(\varepsilon), \widehat{\psi}\_3(\varepsilon,0) = \widehat{\psi}\_{3,0}(\varepsilon), \widehat{\psi}\_4(\varepsilon,0) = \widehat{\psi}\_{4,0}(\varepsilon), \widehat{\psi}\_5(\varepsilon,0) = \widehat{\psi}\_{5,0}(\varepsilon). \tag{45}
$$

As ς = 1 and ψ1 0, ψ2 0, ψ3 0, ψ4 0, ψ5 0 and *<sup>H</sup>*ψ1 0, *<sup>H</sup>*ψ2 0, *<sup>H</sup>*ψ3 0, *<sup>H</sup>*ψ4 0, *<sup>H</sup>*ψ5 0, then Equations (30)–(34) are obtained as

$$
\widehat{\psi}\_1(\varsigma, 1) = \widehat{\psi}\_1(\varsigma), \\
\widehat{\psi}\_2(\varsigma, 1) = \widehat{\psi}\_2(\varsigma), \\
\widehat{\psi}\_3(\varsigma, 1) = \widehat{\psi}\_3(\varsigma), \\
\widehat{\psi}\_4(\varsigma, 1) = \widehat{\psi}\_4(\varsigma), \\
\widehat{\psi}\_5(\varsigma, 1) = \widehat{\psi}\_5(\varsigma), \\
\widehat{\psi}\_6(\varsigma, 1) = \widehat{\psi}\_6(\varsigma), \\
\widehat{\psi}\_7(\varsigma, 1) = \widehat{\psi}\_7(\varsigma), \\
\widehat{\psi}\_8(\varsigma, 1) = \widehat{\psi}\_8(\varsigma), \\
\widehat{\psi}\_9(\varsigma, 1) = \widehat{\psi}\_1(\varsigma), \\
\widehat{\psi}\_{10}(\varsigma, 1) = \widehat{\psi}\_{10}(\varsigma), \\
\widehat{\psi}\_{11}(\varsigma, 1) = \widehat{\psi}\_{11}(\varsigma), \\
\widehat{\psi}\_{12}(\varsigma, 1) = \widehat{\psi}\_2(\varsigma), \\
\widehat{\psi}\_{13}(\varsigma, 1) = \widehat{\psi}\_3(\varsigma), \\
\widehat{\psi}\_{14}(\varsigma, 1) = \widehat{\psi}\_4(\varsigma), \\
\widehat{\psi}\_{15}(\varsigma, 1) = \widehat{\psi}\_5(\varsigma), \\
\widehat{\psi}\_{16}(\varsigma, 1) = \widehat{\psi}\_6(\varsigma), \\
\widehat{\psi}\_{17}(\varsigma) = \widehat{\psi}\_7(\varsigma), \\
\widehat{\psi}\_{18}(\varsigma) = \widehat{\psi}\_2(\varsigma), \\
\widehat{\psi}\_{19}(\varsigma) = \widehat{\psi}\_2(\varsigma), \\
\widehat{\psi}\_{10}(\varsigma) = \widehat{\psi}\_2(\varsigma), \\
\widehat{\psi}\_{11}(\varsigma) = \widehat{\psi}\_2(\varsigma), \\
\widehat{\psi}\_{12}(\varsigma) = \widehat$$

In order to ge<sup>t</sup> *m*th- order deformation equations, Equations (30)–(34) are differentiated *m*- times with respect to ς, after substituting ς = 0 and dividing both sides by *m*!. Finally, the *m*th- order deformation equations take the following forms:

$$\hbar L\_{\psi\_1}[\psi\_{1,m}(\varepsilon) - \chi\_m \psi\_{1,m-1}(\varepsilon)] = \hbar\_{\psi\_1} R\_{1,m}(\varepsilon). \tag{46}$$

$$L\_{\psi\_2}[\psi\_{2,m}(\boldsymbol{\varsigma}) - \chi\_m \psi\_{2,m-1}(\boldsymbol{\varsigma})] = \hbar\_{\psi\_2} R\_{3,m}(\boldsymbol{\varsigma}).\tag{47}$$

$$\left\| L\_{\psi 3} \right\| \psi\_{3,m}(\boldsymbol{\zeta}) - \chi\_{\psi 3} h\_{\psi 3} - 1(\boldsymbol{\zeta}) \right\| = \hbar\_{\psi 3} R\_{3,m}(\boldsymbol{\zeta}).\tag{48}$$

$$L\psi\_4\left[\psi\_{4,m}(\boldsymbol{\varepsilon}) - \chi\_{\psi\_4}h\_{\psi\_4-1}(\boldsymbol{\varepsilon})\right] = \hbar\psi\_4 R\_{4,m}(\boldsymbol{\varepsilon}).\tag{49}$$

$$L\_{\psi\_{\mathbb{S}}}[\psi\_{\mathbb{S},m}(\boldsymbol{\varepsilon}) - \chi\_{\psi\_{\mathbb{S}}}h\_{\psi\_{\mathbb{S}}-1}(\boldsymbol{\varepsilon})] = h\_{\psi\_{\mathbb{S}}}R\_{\mathbb{S},m}(\boldsymbol{\varepsilon}).\tag{50}$$

with boundary conditions

$$
\widehat{\psi}\_{1,m}(1) = -\mu\_2 \widehat{\psi}\_{1,m}^{'}(1), \widehat{\psi}\_{1,m}(0) - 1 = \mu\_1 \widehat{\psi}\_{1,m}^{'}(0) \tag{51}
$$

$$
\widehat{\psi}\_{2,m}(1) = -\mu\_2 \widehat{\psi}\_{2,m}^{'}(1), \widehat{\psi}\_{2,m}(0) - 1 = \mu\_1 \widehat{\psi}\_{2,m}^{'}(0) \tag{52}
$$

$$
\widehat{\psi}\_{3,\mathfrak{m}}^{\\\\\uparrow}(0) = \mu\_1 \widehat{\psi}\_{3,\mathfrak{m}}^{\\\uparrow}(0) = \widehat{\psi}\_{3,\mathfrak{m}}(0) = 0,\\
\widehat{\psi}\_{3,\mathfrak{m}}(1) = 0,\\
\widehat{\psi}\_{3,\mathfrak{m}}^{\\\uparrow}(1) = -\mu\_2 \widehat{\psi}\_{3,\mathfrak{m}}^{\\\uparrow}(1) \tag{53}
$$

$$
\widehat{\psi}\_{4,m}^{\\'}(0) = \mu\_1 \widehat{\psi}\_{4,m}^{\\'/\prime}(0) = \widehat{\psi}\_{4,m}(0) = 0,\\
\widehat{\psi}\_{4,m}(1) = 0,\\
\widehat{\psi}\_{4,m}^{\\'/\prime}(1) = -\mu\_2 \widehat{\psi}\_{4,m}^{\\'/\prime}(1) \tag{54}
$$

$$
\widehat{\psi}\_{5,m}(1) = -\mu\_2 \widehat{\psi}\_{5,m}^{'}(1), \widehat{\psi}\_{5,m}(0) - 1 = \mu\_1 \widehat{\psi}\_{5,m}^{'}(0) \tag{55}
$$

where

$$R\_{1,m}(\boldsymbol{\zeta}) = \boldsymbol{\psi}\_{1,m-1}^{\prime \prime} - R \sum\_{k=0}^{m-1} \boldsymbol{\psi}\_{1,m-1} \boldsymbol{\psi}\_{3,k}^{\prime} + R \sum\_{k=0}^{m-1} \boldsymbol{\psi}\_{3,m-1} \boldsymbol{\psi}\_{1,k}^{\prime} \tag{56}$$

$$R\_{2,m}(\boldsymbol{\varsigma}) = \psi\_{2,m-1}^{\prime \prime} + R \sum\_{k=0}^{m-1} \psi\_{3,m-1} \psi\_{2,k}^{\prime} \tag{57}$$

*Mathematics* **2019**, *7*, 748

$$R\_{3,m}(\boldsymbol{\varsigma}) = \psi\_{3,m-1}^{\prime//\prime//} - R \sum\_{k=0}^{m-1} \psi\_{3,m-1}^{\prime} \psi\_{3,k}^{\prime//} + R \sum\_{k=0}^{m-1} \psi\_{3,m-1} \psi\_{3,k}^{\prime//\prime//} \tag{58}$$

$$R\_{4,m}(\boldsymbol{\varepsilon}) = \psi\_{4,m-1}^{\prime \prime \prime \prime \prime} - 2R \sum\_{k=0}^{m-1} \psi\_{4,m-1} \psi\_{4,k}^{\prime \prime \prime \prime} - M^2 \sum\_{k=0}^{m-1} \psi\_{4,k}^{\prime} \tag{59}$$

$$R\_{5,m}(\boldsymbol{\varphi}) = \psi\_{5,m-1}^{\prime \prime \prime} - R \sum\_{k=0}^{m-1} \psi\_{5,m-1} \psi\_{4,k}^{\prime} - 2 \sum\_{k=0}^{m-1} \psi\_{4,m-1} \psi\_{5,k}^{\prime \prime \prime} - M^2 \psi\_5 \tag{60}$$

A well-known software called MATHEMATICA has been used to solve the modeled problem.
